Question

In $△ABC,XY||BC$ and XY divides the triangle into two parts of equal areas. Find $\left(\frac{BX}{AB}\right)$.

Answer 1

In $△ABC$:

$XY||BC,ar\left(△ABC\right)=2\times ar\left(△AXY\right)$

$△ABC\sim △AXY$ $[\because XY||BC]$

$\frac{ar\left(△ABC\right)}{ar\left(△AXY\right)}=\frac{A{B}^2}{A{X}^2}$ [$\because$ Theorem of area of similar triangles]

$\frac{2}{1}=\frac{A{B}^2}{A{X}^2}=>\frac{\sqrt{2}}{1}=\frac{AB}{AX}=>\frac{1}{\sqrt{2}}=\frac{AX}{AB}=>1-\frac{1}{\sqrt{2}}=1-\frac{AX}{AB}=>\frac{\sqrt{2}-1}{\sqrt{2}}=\frac{AB-AX}{AB}$

$\therefore \frac{\sqrt{2}-1}{\sqrt{2}}=\frac{BX}{AB}$

$\therefore \frac{BX}{AB}=\frac{2-\sqrt{2}}{2}$